trigonometric help 1!!!!!!!!!!!!!!!!

Question

If sinA= -12/13, Q3, and cosB=4/5, Q1, calculate the exact value:

a) sec (A-B)

b) cot (A-B)

Answer 1

The given trigonometric fuctions values are sinA = -12/13, Q3, and cosB = 4/5.

Apply pythagorean identity: c²= a²+ b²

sinA  = a/c =  - 12/13, So a = - 12, b = 13

b²= 160 - 144 = 15

b = 5

cosA = b/c = -5/13, So b = -5 , c = 13

cosB = b/c = 4/5, b = 4 , c = 5

c²= a²+ b²

a²= 9 ⇒ a = 3

sinB = a/c = 3/5, a = 3 , c = 5

a) sec ( A -B ) = 1/cos( A -B)

= 1/ cosA cosB +sinA sinB

= 1/ ( -5/13) (4/5) + ( -12/13) (3/5)

= 1/ ( -4/13) +[ ( -60+39)/65]

= 1/ ( -41/5)  = - 5/41.

Answer 2

The given trigonometric fuctions values are sinA = -12/13, Q3, and cosB = 4/5 Q1.

Apply pythagorean identity: c²= a²+ b²

sinA  = a/c =  - 12/13, So a = - 12, b = 13

b²= 160 - 144 = 15

b = 5

cosA = b/c = -5/13, So b = -5 , c = 13

cosB = b/c = 4/5, b = 4 , c = 5

c²= a²+ b²

a²= 9 ⇒ a = 3

sinB = a/c = 3/5, a = 3 , c = 5

b) cot ( A -B )  = cos (A -B)/ sin (A -B)

cos (A -B) = cosA cosB +sinA sinB

=  ( -5/13) (4/5) + ( -12/13) (3/5)

= ( -4/13) +[ ( -60+39)/65]

=   -41/5

sin (A -B) = sinA cosB - cosA sinB

= ( -12/13) (4/5) - ( -5/13) (3/5)

= -8/65 +3/13 = 7/5

Therefore cot (A-B) = ( -41/5)./(7/5)

= -41/7.